Capt Ayhab

Welcome to "free-basing", Irandokht

Irandokhtar khanom, your response of interpreting 12 in base 12 is perfectly valid. In fact, you don't even have to resort to base 10, since 7 is written like that in any base greater than 7, including base 12. Dash Kapitan is now changing the question. His initial wording was "half of 12" so it's legit to interpret "12" in any mathematically acceptable way. If he'd spelt out "twelve" it may have been different. I thought this may have been a word puzzle so I skipped it. But all I can say is, "you complete me!"

Captain, your statements about independent and identical coin tosses are true, but how does that negate Jaleh's arguments? All she was saying is that when one player has only one coin more than the other, the answer is simply 1/2. But when there's 2 or more coins differential then the answer depends on the number of coins you gave the players. She gave some simple examples of that. Her probability arguments seem sound to me, unlike her pop psychology which is foote ab.

IRAN khanom dokht

Base is decimal[10]  will that help?

As too what i wrote regarding coin toss. I am not making it up, you can see that in any and every elementary probability concepts. I do not doubt Ms. Jaleho's credentials by no means.

Probability of coin toss is [simple] and [independent] hence 1/2 head, 1/2 tail

-YT

dear Capt'n

I don't know Jaleho personally but from what I have seen, I wouldn't be surprised if she'd be a professor herself, she's quite a smart lady.

as for half of 12=7 I have to pull a maghshoosh here and say half of 12 in base 12 (14) would be 7 in base 10

so your equation is correct, but the base is missing. 

IRANdokht

hala

How about:

Half of 12 = 7 ??????????

-YT

Jaleho

I did not want to pull rank on you by solving the problem, since grad level stat is one of the course I teach, but I think you guys are totally off on the coin deal.

Coin toss is a perfect SIMPLE probability problem. Meaning it is Simple[Mutually Exclusive Events] because there are 2 and only 2 possible outcome[Collectively Exhaustive Events]. They are also independent of one another, meaning outcome of one toss does not affect the outcome of the other coins.

with that in mind, a coin toss has the formula of: outcome/total possible outcomes. Simply put each coin has 1/2, or 50% probability of landing a Head or a Tail. Now since one coin toss does not affect the probability of the other, then each and every coin has the same probability.

whether having one coin or 10 coins, probability does not change, as a result. there are SAME probability of SHIR in both of our tosses. this is the statistical calculation.

However i think your problem had a little bit of trick that being the one who has 2 extra coins get MORE heads[so to speak] with SAME PROBABILITY.

-YT

Maghshoosh jon, really, I truly

appreciated that you made sure that there is not any other solution to be found in the cranny and nooks of your house or mine. As you know, it is about time for "khaneh takani," so it would be awful to walk around and think that you might be able to move around a digit in :

101-102=1 in other forms and find something, like yet another meaningful solution like:

101-10=21  or  110-102=1

Of course, for most of us the above two are as meaningless as the original 101-102=1. But, in Maghshoosh's mind they immediately turn into

10-3=7    and   12-11=1

So, I totally understand your urge in proving that there's no other base but three to clog up your brain  unnecessarily :-)

It was awesome, thank!

Avval Irandokht,

OK deary, the reason I threw one more coin was to show the uniqueness of my just one extra coin case. Then for illustration, I assumed that Captain has 2 more coins than Majid, to make you happy too.

So, let's say K has 2 coins and M has none. When they throw their coins, the possible outcomes was only that of Kaptain's coins:

HH, HT, TH, TT. In this case the probability of K having more heads than M is 3 out of 4.

Then I said that you can try K having 3 coins and M having 1 coin, still K has two more coins. Now for these four coins we have 2*2*2*2 or 16 possible outcomes. Out of these, there is one case that K can have less heads (TTT and H); 4 case that they have equal heads (TTT and T) and another 3 cases of (_ _ _ and H) where M's first or second or third coin is a H. So, you have 1/16 for less head, 4/16 for equal heads, and remaining 11/16 for more heads.

So, in the first case the probability of K having more heads was 3/4 whereas in the second case it was 11/16. That is, it depends on how many coins they have, even when K has just 2 more coins than M.

But, if K has just one more head than M, regardless of how many coins they have, the probability of M having more heads than K is always 1/2 because K has either more heads or more tails than M, but not both!

vel nemikonand

Irandokht, I made the mistake of checking this page and Kapitan va Jaleh hey ab nabat mizarand jeloye bacheh. Your suggestion w/ the 3s means that you don't need the factorial to do it w/ the 9s. digeh nemidoonam nesfe 12 misheh 7 cheh sigheie, nemikham ham bedoonam.

too fast...lol

i was just flexing my fingers all ready to go on the 5th grade question and suddenly maghshoosh is there with the answer! 

majid is busy packing and moving... he can't come out to play just yet.  :-0

?

Half of 12 = 7 ?

-YT

maghshoosh you're too fast

saram geej raft baba! as soon as I saw Capt's math problem, your answer was there. (show off) :-P    (just kidding)

ok your answers are really good, I only have a different suggestion for 3 

3x3-3=6

nemizarin adam be kar o zendegish bereseh

:o))

IRANdokht

The return of the Captain.

Captain, you continue to sow corruption in the minds of our impressionable youth and we are helpless to stop you. Irandokht bargard, the tooth fairy in the form of Kapitan is granting your wishes.

Since I don't know how to include math symbols here, some made-up notation. S(n) would mean n under the square-root symbol. C(n) would mean n under the cube-root symbol, which is the square-root symbol w/ a small 3 on the upper left corner of it. And for an integer n, recall the factorial n! which means n multiplied by all the non-zero positive integers smaller than it, so that 3!=3x2=6.

(1+1+1)!=6
2+2+2=6
3-3+3!=6
S(4)+S(4)+S(4)=6
5/5+5=6
6-6+6=6
7-7/7=6
C(8)+C(8)+C(8)=6
9-9+(S(9))!=6

5th grade Math Problem

By placing appropriate operators(+ - X / etc) prove the equations :

1  1  1   = 6

2  2  2  = 6

3  3  3  = 6

4  4  4  = 6

5  5  5  = 6

6  6  6  = 6

7  7  7  = 6

8  8  8  = 6

9  9  9  = 6

Good Luck ... ;-)

-YT

Jaleho jan did hala! :o)

1/2 or 50%?  6 or half a dozen? ;-) so what's his chance with two extra coins?

I already wrote a faal-e hafez on Mr Yassari's new blog. I just love those don't you?  Javad is always right on time too!

I want to know if Maghshoosh can moshaereh as well as he does math problems, what do you think?  

IRANdokht

istgah avaz shod?

Jaleh khanom, you really finesse those sharp turns. 1st come all the can't-be-maghshoosh assurances, then pop the check-your-vasvas-head diagnosis and the kitchen-drawer crack. Shepeshe maghze man keh panjta malagh zad, dotash varooneh. A school of fish would really feel at home following you.

Puzzle me crazy.

Irandokht, I didn't mean to be discouraging of that thread, just that puzzles can become like khoreh to me, and there's no puzzeloholics anonymous. So to save myself I accused you guys of fanaticism. Also having nerdy tendencies, I was just engaging in silly punning w/ the word "baseless," not meant it literally. BTW, Majid never posted his full 12-ball solution & JD was asking for it, if you wanna post it. Well, it was fun while it lasted; short & sweet.

Irandokht,

you ain't leaving, are you? :-(

Agha Majid will come out of his happy hour in one peice and we'll start again....OR ELSE, Javad agha has another moshaereh thread it seems!

Or Irandokht jan,

a little less maghshoosh solution that I had in mind was:

since M has only one more coin than K, M  either has more heads than K or more tails than K, but not both! Because if he had both more heads and more tails, he would have to have at least two more coins. Because of symmetry then , the probability of his having more heads is 1/2.

Maghshoosh jon, nice expansion on the 101-102=1 question. Great job, but you truly have to have your head checked with your vasvas making sure you have not missed any other solutions! Boy, I was so happy with just one. Are you sure there aren't any more in the kitchen drawer? Look again!

oh noooooo it's over?

Jaleho you just ruined my calculation by throwing in an extra coin once again. My female intuition jaa zad yeho! ;-)

I loved the "shepesh ghap mindazeh" (long time since I heard that one)

and to you my not-so-maghshoosh of a friend, thanks for the support! We're all right, nothing wrong with preferring this discussion over the sensitive topics of ME, nuclear stuff, etc... I knew you could come up with a formula to back up the answers: nicely done! but the conclusion you made of Foaad's clever answer sounded a little bitter my friend, what's the matter? can't accept to share the top prize? ;-)

I also loved our little wet kitty cat writing a whole code for the riddle with nested if statements LOL that was priceless.

Not sure why Professor Ayhab gave up on the riddles, but I sure enjoyed hanging in there and reading all the answers. 

It's been fun and thanks for making me think hard everyone!

IRANdokht

Someone stop this insanity!

And I thought I was being fanatical by checking this thread. At least, you ladies are charmingly "fanatical." ID khanom, I think your answer to Jaleh's coin toss problem is correct, though you may not have stated your reasoning clearly. Probabilities can be counter-intuitive. Suppose Dash Kapitan (K) & Haji Majid (M) each have only 1 coin. What are the odds that M will get more H(eads) than K? That's the odds of K getting T(ail) and M getting H, which is 1/4. Is your reasoning consistent w/ this simple case?

So say K has n coins and M n+1. M sets one of his coins aside. They both throw their n coins. The outcomes can only be: same number of H for both, K gets more H or M gets more H. By symmetry (each threw n identical coins), the last 2 cases have equal probabilities, which we denote by p_1. M will only proceed to throw his extra coin if the 1st case happened, to which we assign probability p_2, in which case it's 50/50 that he'll get 1 more H. So the probability of M getting more H is (p_1+.5*p_2)/(p_1+p_1+p_2)=1/2. So your female intuition was right on!

Now to exceed all bounds of reasonable fanaticism, there's one more solution to the 101-102=1 puzzle. To see if there are other solutions by moving a digit horizontally and considering all bases, we move each digit in all possible positions and solve a quadratic equation to see if there's an n>2 integer (for base n) solution. So in the 1st case, we tried 101-10 vs 21. In base n, 101=n^2+1 (n^2 meaning n-squared), 10=n and 21=2n+1. n^2+1-n=2n+1 only has n=3 as a meaningful solution. Another try would be 110-102=1, which results in (n^2+n)-(n^2+2)=1, i.e. n=3. So there's another ternary solution. I think I considered all permutations of a digit and couldn't find any more solutions in any base, but I could have missed some. BTW Jaleh, Foad's superscript-2 solution is not just base 10. 10=n in base n, so that its square may be written as 100 in any base, so that the Foad solution is satisfied in any base. Hence, we may call that solution "baseless"!

Even Dash Kapitan bowed off stage gracefully. What's wrong w/ the rest of us?

I am Grateful

Allow me to take this opportunity to thank each and everyone of you beautiful ladies and gentleman for commenting on this blog.

Baby Lion:

Sorry for delay in answer, you got it all right.

-YT

it was a nested if, but editor is not showing the indents..

-----x = the randow answerer 

-----Ask y if x is The Random Answerer

-----if  y's answer is YES then

-----------y = The Truth Teller  and  z = The Liar

-----else "y's answer is NO"

------------z = The Truth Teller  and y = The Liar     

-----endif

else  "x's answer is NO"
x either is The Random Answerer  OR  The Truth Teller

Ask y if  y is The Truth Teller

if y's answer is YES then

-----y = The Liar

---- Ask y if x is The Truth Teller

-----if y's answer is  YES  then

---------x = The Random Answerer  and  z = The Truth Teller

-----else  "y's answer is NO"

----------z = The Random Answerer  and x = The Truth Teller

-----endif

else  "y's answer is NO"

------y either is The Random Answerer  OR  The Truth Teller 

------z = The Liar

------Ask z if x is The Truth Teller

------if z's answer is YES then 

-----------because z is a liar then  x = The Random Answerer  

------------and y = The Truth Teller

------else " z's answer is NO"

------------because z is a liar then y = The Random Answerer  

------------- and x = The Truth Teller         

-------endif           

endif

Irandokht jan,

your reasoning is not sound, for example what if agha majid had two more coins than captain?

For the sake of illustration, assume capatain has 2 coins and agha Majid none. (happy now?!) OK, then when they both throw their coins, captain would have one of these possibilities:

HH HT TH TT

while to jibeh agha Majid shepesh seh ghap mindazeh :-) 

That is, the probability of captain having more heads than Majid is 3/4. You can try if one of them had 3 coins and the other had one coin to see how the probability would change. So, the feature I gave, of one having just one more coin than the other, is important in the reasoning. 

It is tricky!!

Puss jan

You are only allowed 3 questions addressed to each of the three (liar, truth-teller, random):)

Also just staying on the initial para, what if the random-teller does not say "Yes" and says "NO" just like the other two!!

capt, who is who capitan!

x = the randow answerer 

Ask y if x is The Random Answerer

if  y's answer is YES then y = The Truth Teller

z = The Liar

else "y's answer is NO"

z = The Truth Teller

y = The Liar     

endif <br>

else  "x's answer is NO"
x either is The Random Answerer  OR  The Truth Teller

Ask y if  y is The Truth Teller

if y's answer is YES then

y = The Liar

  Ask y if x is The Truth Teller

if y's answer is  YES  then

x = The Random Answerer

z = The Truth Teller

else  "y's answer is NO"

z = The Random Answerer

x = The Truth Teller

endif

else  "y's answer is NO"

y either is The Random Answerer  OR  The Truth Teller 

z = The Liar

Ask z if x is The Truth Teller

if z's answer is YES then 

because z is a liar =>    x = The Random Answerer  

y = The Truth Teller

else " z's answer is NO"

because z is a liar =>    y = The Random Answerer  

x = The Truth Teller         

endif           

endif

Jaleh jan I agree

about maghshoosh not showing any sign of eghteshash!  I couldn't figure out the answer even after I read M's comment 3 times!! (lets keep that a secret)

to answer your puzzle:

first may I ask why you gave Capt'n one less coin that Majid? 

second: since the probability of that extra coin being shir is 50%, the coins split equally between them and all coins fair, then the chances of majid getting one extra shir is 50% too.

unless you guys have a whole lot of formula again and my diminishing brain power is missing it.

IRANdokht

PS: of course I am keeping an eye on this thread, watching that little number next to it on most discussed list and I get all excited when there is a new entry! YAY!

yeah... I know... I lead a boring life  :o)

this maghshoosh has

such a bi-eghteshash brain! Any kitty can mess up two hundred yarns in his brain and he seem to untangle them neatly!

OK, here's "shir ya khat" puzzle. Hope people still look at this thread:

Agha Majid has one more coin than Captain. Both throw all of their coins simultaneously and observe the number that come up "shir".  what is the probability that agha Majid gets more "shir" than Captain? Assume all the coins are fair.

capt

after they got the $3  back:

They paid 3 x $9 = $27  = Bellhop $2  + $25 hotel owner

Engineering an Empire - The Persians - Part 1of 5

//www.youtube.com/watch?v=8aukC8GBEsU

//www.youtube.com/watch?v=nSm68TxRBRo

//www.youtube.com/watch?v=N5mwOovpetA

//www.youtube.com/watch?v=qaV42Se3yq4

//www.youtube.com/watch?v=F8H-jYYnkt0

enjoy if you have not seen them by now..

maybe someone from those time can help us ;)

The Kapitan's who's who.

Lets label the men as A, B & C, and refer to the truth-teller, liar and random-answerer as T, L & R, respectively. I can think of a couple of different solutions, but here's one. Ask A, "If I were to ask you if B is R, what would be your answer?" If A is not R, his answer would be the correct one. B/c even if A is L, to answer the question he'd have to negate the false answer he'd give to that hypothetical question. If A is R, then it doesn't matter what he answers b/c B & C are not R. So if the answer to that question is yes then C is not R, and if the answer is no then B is not R. Lets say B is not R. And we continue w/ this sort of trick, but to spell it out ...

Ask B, "If I were to ask you if A is R, what would be your answer?" Since B is not R, even if he is L, if he says yes then A is R, otherwise C is R. Lets say we now know that A is R. Then ask B, "If I were to ask you if C is T, what would be your answer?" That answer would again be the correct one. B/c if B is T, then he'd say no, and if B is L he'd say yes (b/c of the double negative), and these would be the correct identifications.